Question: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 + x}{x + 5} = \dfrac{-8x - 20}{x + 5}$
Solution: Multiply both sides by $x + 5$ $ \dfrac{x^2 + x}{x + 5} (x + 5) = \dfrac{-8x - 20}{x + 5} (x + 5)$ $ x^2 + x = -8x - 20$ Subtract $-8x - 20$ from both sides: $ x^2 + x - (-8x - 20) = -8x - 20 - (-8x - 20)$ $ x^2 + x + 8x + 20 = 0$ $ x^2 + 9x + 20 = 0$ Factor the expression: $ (x + 4)(x + 5) = 0$ Therefore $x = -4$ or $x = -5$ However, the original expression is undefined when $x = -5$. Therefore, the only solution is $x = -4$.